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An Idea that is always used by my professor that I do not know how to prove it:

If $X$ is topological space and $x\in X$ and $\{x\} \times I \subset V$ where $V$ is open in $X \times I,$ why we are always sure that there exists a nhbd $U$ of $x$ in $X$ such that $\{x\} \times I \subset U \times I \subset V.$

Is there a prove for this fact, specifically, we do not have any extra condition on our topological space.

My idea is:

1-Since $V$ is open, then every point in $V$ (including $x$) has a nhbd, say $B(r; x)$ lying in $V.$ We are sure that there exists $B(r/2; x)$ lying in a smaller neighborhood of $V$ call it $U.$

Is this a proof or I should add more stuff? or is it wrong?

2- What if we replace $I$ the unit interval by the real line $\mathbb R$?

Intuition
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  • Your proof is wrong on two points. The first one: How do you know that your space is metrizable? (In general false The second: how do you know that halving the radius results in a strictly smaller neighborhood? (In general, it does not.) – Moishe Kohan Jan 20 '23 at 22:06
  • Third: $x\notin V$ and $B(r;x)\subset X\not\subset X\times I.$ Fourth: shrinking (Moishe's second point) is anyway useless. – Anne Bauval Jan 20 '23 at 22:18
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    See tube lemma. Btw you should make explicit at the beginning of your post that $I=[0,1].$ – Anne Bauval Jan 20 '23 at 22:35
  • @MoisheKohan do we need the space to be metrizable to speak about the notion of a radius? – Intuition Jan 21 '23 at 05:47
  • @AnneBauval are you speaking about that the nbhd must be an ordered pair? I did not get your point in third. – Intuition Jan 21 '23 at 05:50
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    @MoisheKohan I did not get your second point. – Intuition Jan 21 '23 at 05:54
  • @Secretly: "do we need the space to be metrizable to speak about the notion of a radius?" Yes; look up the definition of $B(r;x)$. – joriki Jan 21 '23 at 07:30
  • @Secretly: 1. Ok, then define what $B(r; x)$ is without a metric. 2. Consider a metric space $X$ with the discrete metric, $d(x,x)=0, d(x,y)=1$ for all $y\ne x$. Then $B(r; x)={x}$ for all $r<1$. In particular, $B(r; x)=B(r/2;x)$ for all $x\in X$, all $r<1$. – Moishe Kohan Jan 21 '23 at 08:20
  • @Secretly I am pointing the fact that you are given $x\in X$ and $V\subset X\times I$ and in your 1., you pretend $x\in V$ and $B(r;x)\in V.$ – Anne Bauval Jan 21 '23 at 08:51
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    I think we answered your question 1 by pointing various mistakes in your attempt. For a proof, see for instance the link I above ("tube lemma"), or that post: Doubt about proof in Tube Lemma and linked or related posts. The answer to your question 2 is: no because $\Bbb R$ is not compact. https://math.stackexchange.com/questions/4353931/the-importance-of-compactness-in-the-tube-lemma – Anne Bauval Jan 21 '23 at 08:53

1 Answers1

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As Anne Bauval suggests in the comments, you should look up the tube lemma, which is the desired statement with the unit interval $I$ replaced more generally with any compact space $Y$.

I want to address your second question, which provides some intuition for why we want compactness for the space $Y$ we are producting with.

Let $X = \mathbb{R}$, and consider $V \subset \mathbb{R} \times \mathbb{R}$ defined by $$V = \{(x, y) \mid -e^{y} < x < e^{y}\}.$$ Check for yourself that this is an open set.

Consider $\{0\} \times \mathbb{R} \subset V.$ We claim there doesn't exist an open set $U \subset \mathbb{R}$ containing $\{0\}$ such that $$\{0\} \times \mathbb{R} \subset U \times I \subset V.$$ To see this, note that any open set $U$ of $\mathbb{R}$ containing $\{0\}$ must contain $(-\varepsilon, \varepsilon)$ for some $\varepsilon > 0$. This means that $V$ must contain $(-\varepsilon, \varepsilon) \times \mathbb{R}$. But for small enough $y$ we have $e^{y} < \varepsilon$, so this is impossible!

So the claim doesn't hold when $I$ is replaced with $\mathbb{R}$.

ckefa
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